I've been doing questions regarding the shortest distance between lines/planes and points , and I've come across a question asking to find the shortest distance between a line and a plane which are both parallel to each other.
Now I haven't seen this type of question before so I drew a diagram to work out the shortest distance which I believed to be the perpendicular distance between the line and the plane.
In the picture $R$ is a point on L such that $\vec{BR}.\vec{AR} = 0$ and $A$ and $B$ are known points on the line and plane.
The equations for the plane and line are given as:
$\prod r = \begin{pmatrix}
1\\
3\\
4\\
\end{pmatrix}
+ u\begin{pmatrix}
4\\
1\\
2\\
\end{pmatrix}
+
v\begin{pmatrix}
3\\
2\\
-1\\
\end{pmatrix}
$
$L\: r = \begin{pmatrix}
2\\
1\\
-3\\
\end{pmatrix}
+t \begin{pmatrix}
2\\
3\\
-4\\
\end{pmatrix}
$
Where $\vec{OR} = \begin{pmatrix}
2 + 2t\\
1 + 3t\\
-(3 + 4t)\\
\end{pmatrix}
$
So I found $\vec{BR}$ and $\vec{AR}$ and dotted them together to solve for t where I ended up with this:
$29t^{2} + 24t = 0
\\
t(29t + 24) = 0
\\
t = 0\:\:\:t = -\frac{24}{29}$
So when I subbed the value of t $-\frac{24}{29}$ back into $\vec{AR}$ and found the modulus I was greeted with the anwser of $\sqrt{\frac{990}{29}}$ which is wrong because the correct anwser is $2\sqrt{6}$.
I reworked my answer 3 times but arrived at the same answer everything hence I can only conclude that my logic in my diagram is flawed. If my answer is not the shortest distance then what exactly did I just work out? Can someone please tell me why my method doesn't work?
Thank you
Best Answer
The flaw is in your diagram. For your method to work, the point $A$ must lie on the projection of the line onto the plane, otherwise the vector $\vec{AR}$ will be oblique to the plane.
While drawing the figure you have assumed that the point $A$ is lying on the projection of $L$ onto $\Pi$ while it's actually not.